The motion of a particle is described by the equation `x = a+bt^(2)` where `a = 15` cm and `b = 3 cm//s^2`. Its instantaneous velocity at time 3 sec will be

Options

A. `33 cm s^(-1)`
B. `18 cm s^(-1)`
C. `16 cm s^(-1)`
D. `32 cm s^(-1)`

Explanation

To find the instantaneous velocity of the particle at time \( t = 3 \) seconds, we start with the given position equation:

\[ x = a + bt^2 \]

where \( a = 15 \) cm and \( b = 3 \) cm/s².

### Step 1: Differentiate the position equation with respect to time
The instantaneous velocity \( v \) is given by the derivative of the position \( x \) with respect to time \( t \):

\[ v = \frac{dx}{dt} \]

### Step 2: Differentiate the equation
Now, we differentiate the equation \( x = a + bt^2 \):

- The derivative of \( a \) (a constant) with respect to \( t \) is \( 0 \).
- The derivative of \( bt^2 \) with respect to \( t \) is \( 2bt \).

Thus, we have:

\[ v = 0 + 2bt = 2bt \]

### Step 3: Substitute the values of \( b \) and \( t \)
Now we substitute \( b = 3 \) cm/s² and \( t = 3 \) s into the equation for velocity:

\[ v = 2 \cdot 3 \cdot 3 \]

### Step 4: Calculate the instantaneous velocity
Calculating the above expression gives:

\[ v = 2 \cdot 3 \cdot 3 = 18 \text{ cm/s} \]

### Conclusion
The instantaneous velocity of the particle at \( t = 3 \) seconds is:

\[ v = 18 \text{ cm/s} \]